Предмет: Алгебра, автор: dove7549

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Автор ответа: Miroslava227

$\sin {}^{2} ( \alpha ) + \cos {}^{2} ( \alpha ) + ctg {}^{2} (\alpha ) = 1 + {ctg}^{2} ( \alpha ) = \frac{1}{ \sin {}^{2} ( \alpha ) } \\$

$\cos {}^{2} ( \alpha ) \times (1 + {tg}^{2} \alpha ) = \cos {}^{2} ( \alpha ) \times \frac{1}{ \cos {}^{2} ( \alpha ) } = 1 \\$

$1 - \frac{1}{ \sin {}^{2} ( \alpha ) } = \frac{ \sin {}^{2} ( \alpha ) - 1}{ \sin {}^{2} ( \alpha ) } = - \frac{ \cos {}^{2} ( \alpha ) }{ \sin {}^{2} ( \alpha ) } = - {ctg}^{2} \alpha \\$

$4 - tg \alpha \times ctg \alpha = 4 - 1 = 3$

$\cos {}^{2} ( \beta ) - \cos {}^{2} ( \beta ) \sin {}^{2} ( \beta ) = \cos {}^{2} ( \beta ) (1 - \sin {}^{2} ( \beta )) = \\ = \cos {}^{2} ( \beta ) \times \cos {}^{2} ( \beta ) = \cos {}^{4} ( \beta )$

$\sin {}^{4} ( \beta ) + \sin {}^{2} ( \beta ) \cos {}^{2} ( \beta ) = \\ = \sin {}^{2} ( \beta ) ( \sin {}^{2} ( \beta ) + \cos {}^{2} ( \beta )) = \\ = \sin {}^{2} ( \beta ) \times 1 = \sin {}^{2} ( \beta )$

${tg}^{2} \alpha \times {ctg}^{2} \beta - \sin {}^{2} ( \beta ) = \\ = 1 - \sin {}^{2} ( \beta ) = \cos {}^{2} ( \beta )$

$\frac{1 - \cos {}^{2} ( \beta ) }{ \sin {}^{2} ( \beta ) - 1 } = \frac{ \sin {}^{2} ( \beta ) }{ - \cos {}^{2} ( \beta ) } \times - {tg}^{2} \beta \\$

$\frac{ \cos( \alpha ) + ctg\alpha }{1 + \sin( \alpha ) } = \frac{ \cos( \alpha ) + \frac{ \cos( \alpha ) }{ \sin( \alpha ) } }{ 1 + \sin( \alpha ) } = \\ = \frac{1}{1 + \sin( \alpha ) } \times \frac{ \cos( \alpha )(1 + \sin( \alpha )) }{ \sin( \alpha ) } = \\ = \frac{ \cos( \alpha ) }{ \sin( \alpha ) } = ctg( \alpha )$

10.

$(1 - \cos {}^{2} ( \alpha )) (1 + {tg}^{2} \alpha ) = \\ = \sin {}^{2} ( \alpha ) \times \frac{1}{ \cos {}^{2} ( \alpha ) } = {tg}^{2} \alpha$

11.

$\frac{ \sin( \alpha ) }{ 1 + \cos( \alpha ) } + \frac{1 + \cos( \alpha ) }{ \sin( \alpha ) } = \\ = \frac{ \sin {}^{2} ( \alpha ) + {(1 + \cos( \alpha )) }^{2} }{ \sin( \alpha ) (1 + \cos( \alpha )) } = \\ = \frac{ \sin {}^{2} ( \alpha ) + 1 + 2 \cos( \alpha ) + \cos {}^{2} ( \alpha ) }{ \sin( \alpha ) (1 + \cos( \alpha )) } = \\ = \frac{2 + 2 \cos( \alpha ) }{ \sin( \alpha )(1 + \cos( \alpha ) ) } = \frac{2(1 + \cos( \alpha )) }{ \sin( \alpha )( \cos( \alpha ) + 1)} = \\ = \frac{2}{ \sin( \alpha ) }$

12.

$\frac{ \cos( \alpha ) }{ 1 + \sin( \alpha ) } - \frac{ \cos( \alpha ) }{ 1 - \sin( \alpha ) } = \\ = \frac{ \cos( \alpha ) (1 - \sin( \alpha )) - \cos( \alpha ) (1 + \sin( \alpha )) }{ {1 - \sin {}^{2} ( \alpha ) }^{} } = \\ = \frac{ \cos( \alpha ) (1 - \sin( \alpha) - 1 - \sin( \alpha ) }{ \cos {}^{2} ( \alpha ) } = \\ = - 2 \frac{ \sin( \alpha ) }{ \cos( \alpha ) } = - 2tg \alpha$

13.

$\frac{1 - 2 \sin( \alpha ) \cos( \alpha ) }{ \sin( \alpha ) - \cos( \alpha ) } = \\ = \frac{ \sin {}^{2} ( \alpha ) + \cos {}^{2} ( \alpha ) - 2 \sin( \alpha ) \cos( \alpha ) }{ \sin( \alpha ) - \cos( \alpha ) } = \\ = \frac{( \sin( \alpha ) - \cos( \alpha )) {}^{2} }{ \sin( \alpha ) - \cos( \alpha ) } = \sin( \alpha ) - \cos( \alpha )$

14.

$\sin {}^{4} ( \alpha ) - \cos {}^{4} ( \alpha ) + \cos {}^{2} ( \alpha ) = \\ = ( \sin {}^{2} ( \alpha ) - \cos {}^{2} ( \alpha ) )(\sin {}^{2} ( \alpha ) + \cos {}^{2} ( \alpha )) + \cos {}^{2} ( \alpha ) = \\ = ( \sin {}^{2} ( \alpha ) - \cos {}^{2} ( \alpha )) \times 1 + \cos {}^{2} ( \alpha ) = \sin {}^{2} ( \alpha )$

15.

$\frac{1 + {ctg}^{4} \alpha }{ {ctg}^{2} \alpha + {tg}^{2} \alpha } = (1 + {ctg}^{4} \alpha ) \times \frac{1}{ {ctg}^{2} \alpha + \frac{1}{ {ctg}^{2} \alpha } } = \\ = (1 + {ctg}^{4} \alpha ) \times \frac{ {ctg}^{2} \alpha }{1 + {ctg}^{4} \alpha } = {ctg}^{2} \alpha$

16.

$\frac{ \cos {}^{2} ( \alpha ) }{1 + {tg}^{2} \alpha } - \frac{ \sin {}^{2} ( \alpha ) }{1 + {ctg}^{2} \alpha } = \\ = \frac{ \cos {}^{2} ( \alpha ) }{ \frac{1}{ \cos {}^{2} ( \alpha ) } } - \frac{ \sin {}^{2} ( \alpha ) }{ \frac{1}{ \sin {}^{2} ( \alpha ) } } = \\ = \cos {}^{4} ( \alpha ) - \sin {}^{4} ( \alpha ) = \\ = ( \cos {}^{2} ( \alpha ) - \sin {}^{2} ( \alpha ))( \cos {}^{2} ( \alpha ) + \sin {}^{2} ( \alpha ) ) = \\ = \cos(2 \alpha )$